# Am I permitted to use the truth of the base case during the inductive step in a proof using weak induction?

The principle of mathematical induction says that if $S(n)$ is some mathematical assertion about the natural number $n$ such that if $S(1)$ is true, and if the truth of $S(n)$ implies the truth of $S(n+1)$, then $S(n)$ is true for every natural number $n$. Some books refer to this principle as weak induction.

Strong induction allows me to assume that $S(\, j \,)$ is true for all natural numbers $1 \leq j \leq n$; however, the book I am using does not mention it, so I assume that I cannot use it.

Question In general, is it permissible to use the truth of the base case during the inductive step when employing weak induction?

Here is the particular problem I am working on, which comes from Fitzpatrick's Advanced Calculus text.

I have to prove that if $n$ is a natural number, and $S(m)$ is the statement concerning the natural number $m$, "If m < n, then $n-m \in \mathbb{N}$". During the inductive step, I assumed that $m+1 < n$. I then reasoned that if I could show that $n-1$ is a natural number, then the proof follows immediately. However, in order to show that $n-1$ is a natural number, I demonstrated that if $n>m+1$, then $n>1$, which from the base case implies $n-1 \in \mathbb{N}$.

• Yes, of course. – Andrés E. Caicedo Jun 6 '18 at 14:28
• Yes, of course you can, since it has been established at the initial step: it's not a hypothesis. – Bernard Jun 6 '18 at 14:28
• Thanks, guys. I can't believe I never realized that I could do that even though I've been using induction for years haha. – Benedict Voltaire Jun 6 '18 at 14:29
• Btw. You can use strong induction in disguise by proving with weak induction the assertion $T(n):=S(k)\text{ is true for every }k\leq n$ – drhab Jun 6 '18 at 14:31